cho da thuc f(x) voi cac he so nguyen thoa man f(3)-f(4)=5. chung minh f(x)-6 khong co nghiem nguyen
cmr khong co da thuc f(x) nao he so nguyen co the co gia tri f(7)=5 va f(15)=9
Đặt \(f\left(x\right)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0\)\(\left(a_i\in Z\right)\)
Ta có: \(f\left(15\right)=a_n.15^n+a_{n-1}.15^{n-1}+...+a_1.15+a_0=9\)
\(f\left(7\right)=a_n.7^n+...+a_1.7+a_0=5\)
\(\Rightarrow\left(15^n-7^n\right)a_n+\left(15^{n-1}-7^{n-1}\right).a_{n-1}+...+\left(15-7\right)a_1=9-5\)
Mà \(15^k-7^k=\left(15-7\right)\left(15^{k-1}+15^{k-2}.7+...+15^i.7^{k-1-i}+..+15.7^{k-2}+7^{k-1}\right)=8X_k\)
\(\left(X_K\in Z\right)\)
\(\Rightarrow8X_n.a_n+8X_{n-1}.a_{n-1}+...+8a_1=4\)
\(\Rightarrow X_na_n+X_{n-1}a_{n-1}+...+X_1a_1=\frac{1}{2}\text{ (vô lí do }X_k,\text{ }a_k\in Z\text{)}\)
Vậy không tồn tại đa thức hệ số nguyên thỏa f(7) = 5; f(15) = 9.
cho f(x) =ax*2+bx+c biet f(1) .f(2) .f(0) nguyen .chung minh da thuc f(x) nguyen voi moi x
cho da thuc f( x) = x4+ 2x3 -x - 2
a, phan tich f(x) thanh nhan tu
b, chung minh f(x) chia het cho 6 voi moi x la so nguyen
a)\(f\left(x\right)=x^4+2x^3-x-2\)
\(=x^4+2x^3+x^2-x^2-x-2\)
\(=\left(x^2+x\right)^2-\left(x^2+x\right)-2\)
Đặt \(x^2+x=t\) ta có:
\(=t^2-t-2\)\(=\left(t-2\right)\left(t+1\right)\)
\(=\left(x^2+x-2\right)\left(x^2+x+1\right)\)
\(=\left(x-1\right)\left(x+2\right)\left(x^2+x+1\right)\)
Cho da thuc f(x)=ax2+bx+c ; a,b,c la cac so nguyen Chung minh rang khong xay ra dong thoi f(2016)=2017; f(2018)=2018
a. Xac dinh a de nghiem cua da thuc f(x) = 2x-4 cung la nghiem cua da thuc g(x) = x^2 - ax +2b.
b. Cho f(x) = ax^3 + bx^2 + cx + d, trong do a; b; c; d la hang so va thoa man : b = 3a + c
Chung to rang : f(1) = f(-2)
cho da thuc f(x) co bac 4 thoa man f(1)=f(-1);f(2)=f(-2). Chung minh f(2013)=f(-2013)
cho da thuc f(x)=ax^2+bx+c voi a,b,c la cac so thuc . Biet rang f(0), f(1), f(2) co gia tri nguyen . cmr : 2a, 2b cung co gt nguyen
chof(x)=ax^2+bx+cvoi a b c là các số hữu tỉ thỏa mãn 13a+b+2c=0 cmr f(-2)xf(3),nho hon bang 0
Toan lop 7 ma sao kho the?!!!!! Minh bo tay!
cho ham so 2 bien f(x,y) = x^3 +17x +36y ton tai hay khong so nguyen so nguyen x,y thoa man f(x,y) = 2018^2018
cho ham so 2 bien f(x,y) = x^3 +17x +36y ton tai hay khong so nguyen so nguyen x,y thoa man f(x,y) = 2018^2018